Solutions to Problem Set #1

1. Let Ω = a, b, c be a sample space. Let m(a) = 12,m(b) = 1

3and m(c) = 1

6. Find the

probabilities for all eight subsets of Ω.

The eight subsets are: ∅, a, b, c, a, b, a, c, b, c, a, b, c.

m(∅) = 0

m(a) =1

2, m(b) =

1

3, m(c) =

1

6

m(a, b) =5

6, m(a, c) =

2

3, m(b, c) =

1

2

m(a, b, c) = 1

6. A die is loaded in such a way that the probability of each face turning up is proportionalto the number of dots on that face. (For example, a six is three times as probable as a two.)What is the probability of getting an even number in one throw?

If ω1, · · · , ω6 represent the outcomes of rolling a 1, · · · , 6 on the die (respectively), thenthe following conditions must hold:

(1) m(ωi) ≥ 0, i = 1, ..., 6

(2) m(ω1) + · · ·m(ω6) = 1

(3) m(ω6) = 2 ·m(ω3), etc.

(Remember that the first two conditions must always hold for a distribution function).

In order for all three of these conditions to be satisfied simultaneously, we can take1 + 2 + 3 + 4 + 5 + 6 = 21 to be the denominator for our probability weights. Thenm(ω1) = 1

21,...,m(ω6) = 6

21. (Check for yourself that this choice of values of m(ωi) satisfies

the three conditions above!)

Therefore, P (Even) = P (2, 4, 6 = 221

+ 421

+ 621

= 1221

= 47

= 0.57.

7. Let A and B be events such that P (A ∩ B) = 14, P (Ac) = 1

3, and P (B = 1

2. What is

P (A ∪B)?

Recall Theorem 4 from class: P (A ∪ B) = P (A) + P (B)− P (A ∩ B). We already knowthat P (B) = 1

2and P (A ∩ B) = 1

4, so we just need to find P (A). By Theorem 1 part (5),

we know that P (Ac) = 1 − P (A), which is equivalent to saying that P (A) = 1 − P (Ac).Therefore,

P (A) = 1− P (Ac) = 1− 1

3=

2

3.

1

Solutions to Problem Set #1

Putting everything together, we have

P (A ∪B) = P (A) + P (B)− P (A ∩B) =2

3+

1

2=

1

4=

11

12.

10. For a bill to come before the president of the United States, it must be passed by both theHouse of Representatives and the Senate. Assume that, of the bills presented to these twobodies, 60 percent pass the House, 80 percent pass the Senate, and 90 percent pass at leastone of the two. Calculate the probability that the next bill presented to the two groups willcome before the president.

We can rearrange the equation in Theorem 4 to obtain:

P (H ∩ S) = P (H) + P (S)− P (H ∪ S) = .8 + .6− .9 = .5 =1

2.

31. (Paraphrased for the sake of brevity!) Two students tell a teacher that they didn’t arriveat their exam because of a flat tire. In two separate rooms, the teacher asks each of them toidentify the tire that went flat.

(a) Is the probability that they name the same tire 116

?

No. If we assume that the students were lying then the probability that they say thesame things is 1

4, because out of the 16 possible pairs of answers, 4 of them match (to see

this, make a tree!). So, the probability that they guess the same tire is 14. If we assume

that the students were telling the truth, the probability that they identify the same tire is1 (unless one of the students - the passenger in the car - was too drunk to remember thedetails of the incident, in which case the probability that he guesses correctly is 1

4). Either

way, the probability is better than 116

.

(b) In a survey that consists of guessing which tire went flat on a car, 58% of respondentsguessed right front (RF), 11% guessed left front (LF), 18% guessed right rear (RR) and 13%guessed left rear (LR). What is the probability that two random people guessed the same tire?

The probability that both choose RF is (P (RF ))2 = (0.58)2. The same rationale goes forthe other choices. Thus, P (Two Choose Same Tire)= (.58)2+(.11)2+(.18)2+(.13)2 = .3978.

2